Answer:
ΔS° = 180.5 J/mol.K
Explanation:
Let's consider the following reaction.
4 NH₃(g) + 5 O₂(g) → 4 NO(g) + 6 H₂O(g)
The standard molar entropy of the reaction (ΔS°) can be calculated using the following expression.
ΔS° = ∑np × S°p - ∑nr × S°r
where,
ni are the moles of reactants and products
S°i are the standard molar entropies of reactants and products
ΔS° = 4 mol × S°(NO(g)) + 6 × S°(H₂O(g)) - 4 mol × S°(NH₃(g)) - 5 mol × S°(O₂(g))
ΔS° = 4 mol × 210.8 J/K.mol + 6 × 188.8 j/K.mol - 4 mol × 192.5 J/K.mol - 5 mol × 205.1 J/K.mol
ΔS° = 180.5 J/K
This is the change in the entropy per mole of reaction.
Answer:
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Explanation:
Answer:
the density of the cube is 304 g/ cm^3
Explanation:
Because 1520 ÷ 5 = 304 or 304 × 5 =1520
Answer : The correct option is, (D) 100 times the original content.
Explanation :
As we are given the pH of the solution change. Now we have to calculate the ratio of the hydronium ion concentration at pH = 5 and pH = 3
As we know that,
![pH=-\log [H_3O^+]](https://tex.z-dn.net/?f=pH%3D-%5Clog%20%5BH_3O%5E%2B%5D)
The hydronium ion concentration at pH = 5.
![5=-\log [H_3O^+]](https://tex.z-dn.net/?f=5%3D-%5Clog%20%5BH_3O%5E%2B%5D)
..............(1)
The hydronium ion concentration at pH = 3.
![3=-\log [H_3O^+]](https://tex.z-dn.net/?f=3%3D-%5Clog%20%5BH_3O%5E%2B%5D)
................(2)
By dividing the equation 1 and 2 we get the ratio of the hydronium ion concentration.
![\frac{[H_3O^+]_{original}}{[H_3O^+]_{final}}=\frac{1\times 10^{-5}}{1\times 10^{-3}}=\frac{1}{100}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BH_3O%5E%2B%5D_%7Boriginal%7D%7D%7B%5BH_3O%5E%2B%5D_%7Bfinal%7D%7D%3D%5Cfrac%7B1%5Ctimes%2010%5E%7B-5%7D%7D%7B1%5Ctimes%2010%5E%7B-3%7D%7D%3D%5Cfrac%7B1%7D%7B100%7D)
![100\times [H_3O^+]_{original}=[H_3O^+]_{final}](https://tex.z-dn.net/?f=100%5Ctimes%20%5BH_3O%5E%2B%5D_%7Boriginal%7D%3D%5BH_3O%5E%2B%5D_%7Bfinal%7D)
From this we conclude that when the pH of a solution changes from a pH of 5 to a pH of 3, the hydronium ion concentration is 100 times the original content.
Hence, the correct option is, (D) 100 times the original content.
pH of the solution after 24. 00 ml of the hcl has been added is 12.87
millimoles NaOH = mL x M = 24.00 mL x 0.25 M = 6.00
millimoles HCl = 24.00 mL x 0.10 M = 2.40
total volume = 48.00 mL
.................................NaOH + HCl ==>NaCl + H2O
initial.........................6.00.........0............0.........0
added.....................................2.40............................
change.................... -2.40......-2.40.........+2.40.... +2.40
equilibrium.................3.60.........0..............2.40.......2.40
The NaCl contributes nothing to the pH of the final solution. The pH is determined by the excess of NaOH present. (NaOH) = millimoles/mL = 3.60/48.00 = 0.075 M = (OH^-)
pOH = -log (OH^-). Then
pOH = -log (0.075)
pOH =1.1249
As we know,
pH + pOH = pKw = 14.00
pH=14-pOH
pH=14-1.1249
pH=12.87
<h3>
What is pH?</h3>
pH is a logarithmic measure of an aqueous solution's hydrogen ion concentration. pH = -log[H+], where log is the base 10 logarithm and [H+] is the concentration of hydrogen ions in moles per liter.
The pH of an aqueous solution describes how acidic or basic it is, with a pH less than 7 being acidic and a pH greater than 7 being basic. A pH of 7 is regarded as neutral (e.g., pure water). pH values typically range from 0 to 14, though very strong acids may have a negative pH and very strong bases may have a pH greater than 14.
Learn more about pH:
brainly.com/question/491373
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